Q. An organic compound containing $C , H$ and $O$ has $49.3 \%$ carbon, $6.84 \%$ hydrogen and its vapour density is $73$ . Molecular formula of the compound is
Some Basic Concepts of Chemistry
Solution:
Element
$\%$
Relative number of atom
Simplest ratio
$C$
$49.3\%$
$\frac{49.3}{12}=4.1$
$\frac{4.1}{2.74}=1.5 \times 2=3$
$H$
$6.84$
$\frac{6.84}{1}=6.84$
$\frac{6.84}{2.74}=2.5 \times 2=5$
$O$
$43.86$
$\frac{43.86}{16}=2.74$
$\frac{2.74}{2.74}=1 \times 2=2$
Thus, the empirical formula is $C _{3} H _{5} O _{2}$.
Empirical formula weight
$=3 \times 12+5 \times 1+2 \times 16$
$=36+5+32=73$
Molecular weight of the compound $=2 \times V D$
$=2 \times 73=146$
$n=\frac{\text { mol. } wt .}{\text { empirical formula } wt .}=\frac{146}{73}=2$
Molecular formula $=$ Empirical formula $\times 2$
$=\left( C _{3} H _{5} O _{2}\right) \times 2= C _{6} H _{10} O _{4}$
| Element | $\%$ | Relative number of atom | Simplest ratio |
|---|---|---|---|
| $C$ | $49.3\%$ | $\frac{49.3}{12}=4.1$ | $\frac{4.1}{2.74}=1.5 \times 2=3$ |
| $H$ | $6.84$ | $\frac{6.84}{1}=6.84$ | $\frac{6.84}{2.74}=2.5 \times 2=5$ |
| $O$ | $43.86$ | $\frac{43.86}{16}=2.74$ | $\frac{2.74}{2.74}=1 \times 2=2$ |